Re: Capacitor and Force



On Oct 11, 12:15 am, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx>
wrote:
I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or application.
The theory is pretty straight forward and even if my calulations are off by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

I don't think that the force between two point charges is of much
relevance to the force between the two parallel plates of a capacitor.

An easier way of getting there is to note that the energy stored in a
capacitor is 0,5*C*V^2, or 8 microjoules for a 40nF capacitor.

40nF is quite high for an air-gapped capacitor with circular plates
1cm in diameter - that is only 8*10^-5 square metres. The permitivity
of free space is about 9*10^-12 F/m, so your spacing would be about
0.1 micron.

Energy is force times distance, and doubling the gap to 0.2 micron
would halve the capacitance and halve the stored energy, which gives
an attractive force of the order of ten newtons - about one kilogram
weight - but only over about 0.1 micron. Obviously the force is going
to decrease very rapidly as the gap gets larger ...

Hope this helps.

--
Bill Sloman, Nijmegen

.



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